k[X1 , . . . , Xd ].

Solutions to Problems

Chapter 1

1. The primary ideals are (0) and (pn ), p prime.

2. R/Q ∼ k[y]/(y 2 ), and zero-divisors in this ring are of the form cy + (y 2 ), c ∈ k, so

=

they are nilpotent. Thus Q is primary. Since r(Q) = P = (x, y), Q is P -primary.

√

3. If Q = P0 with P0 prime, then Q = P0 , so by Problem 2, P0 = (x, y). But x ∈ Q

n

and x ∈ P0 for n ≥ 2, so Q = P0 for n ≥ 2. Since y ∈ P0 but y ∈ Q, we have Q = P0

/n n

/

and we reach a contradiction.

4. P is prime since R/P ∼ k[y], an integral domain. Thus P is a prime power and

2

=

2

its radical is the prime ideal P . But it is not primary, because x y = z 2 ∈ P , x ∈

/

2

P , y ∈ P.

/

5. We have I ⊆ P1 © P2 and I ⊆ P1 © Q by de¬nition of the ideals involved. For the

2

reverse inclusions, note that if f (x, y)x = g(x, y)y 2 (or f (x, y)x = g(x, y)y), then

g(x, y) must involve x and f (x, y) must involve y, so f (x, y) is a polynomial multiple

of xy.

Now P1 is prime (because R/P1 ∼ k[y], a domain), hence P1 is P1 -primary. P2 is

=

√ 2

2=

maximal and P2 Q = P2 . Thus P2 and Q are P2 -primary. [See (1.1.1) and

(1.1.2). Note also that the results are consistent with the ¬rst uniqueness theorem.]

6. Let M be the maximal ideal of R, and k = R/M the residue ¬eld. Let Mk =

k —R M = (R/M) —R M ∼ M/MM . Assume M —R N = 0. Then Mk —k Nk =

=

(k—R M )—k (k—R N ) = [(k—R M )—k k]—R N = (k—R M )—R N = k—R (M —R N ) = 0.

Since Mk and Nk are ¬nite-dimensional vector spaces over a ¬eld, one of them must

be 0. [k r —k k s = (k —k k s )r because tensor product commutes with direct sum, and

this equals (k s )r = k rs .] If Mk = 0, then M = MM , so by NAK, M = 0. Similarly,

Nk = 0 implies N = 0.

7. We have Z/nZ —Z Z/mZ ∼ Z/(n, m)Z, which is 0 if n and m are relatively prime.

=

8. (M —R N )S ∼ RS —R (M —R N ) ∼ (RS —R M )—R N ∼ MS —R N ∼ (MS —RS RS )—R N ∼

= = = = =

∼ MS —R NS .

MS —RS (RS —R N ) = S

9. By Problem 8, (M —R N )P ∼ MP —RP NP as RP -modules. Thus P ∈ Supp(M —R N )

/

=

i¬ MP —RP NP = 0. By Problem 6, this happens i¬ MP = 0 or NP = 0, that is,

P ∈ Supp M or P ∈ Supp N .

/ /

1

2

10. The ¬rst assertion follows from (1.6.4) and (1.6.6). Since the preimage of a prime

ideal under a ring homomorphism is prime, the second assertion follows from (1.6.4).

11. Say Pin = 0. Then x ∈ Mi i¬ πi (x) ∈ Pi i¬ πi (xn ) = 0 i¬ xn ∈ Ii , and the result

follows.

12. Since Ii consists of those elements that are 0 in the ith coordinate, the zero ideal is the

intersection of the Ii , and Ii ⊇ ©j=i Ij . By Problem 11, the decomposition is primary.

√

Now Ii ⊆ Ii = Mi , and Ii + Ij = R for i = j. Thus Mi + Mj = R, so the Mi are

distinct and the decomposition is reduced.

13. By Problem 12, the Mi are distinct and hence minimal. By the second uniqueness

theorem (1.4.5), the Ii are unique (for a given R). Since Ri ∼ R/Ii , the Ri are unique

=

up to isomorphism.

14. By (1.6.9), the length lRP (MP ) will be ¬nite i¬ every element of APRP (MP ) is

maximal. Now RP is a local ring with maximal ideal P RP . By the bijection of

(1.4.2), lRP (MP ) < ∞ i¬ there is no Q ∈ AP(M ) such that Q ‚ P . By hypothesis,

P ∈ Supp M , so by (1.5.8), P contains some P ∈ AP(M ), and under the assumption

that lRP (MP ) is ¬nite, P must coincide with P . The result follows.

Chapter 2

1. Let Q1 = (2 + i), Q2 = (2 ’ i). An integer divisible by 2 + i must also be divisible by

the complex conjugate 2 ’ i, hence divisible by (2 + i)(2 ’ i) = 5. Thus Q1 © Z = (5),

and similarly Q2 © Z = (5).

2. We have x2 = y 3 , hence (x/y)2 = y. Thus ±2 ’ y = 0, so ± is integral over R. If

± ∈ R, then ± = x/y = f (x, y) for some polynomial f in two variables with coe¬cients

in k, Thus x = yf (x, y). Written out longhand, this is X + I = Y f (X, Y ) + I, and

consequently X ’ Y f (X, Y ) ∈ I = (X 2 , Y 3 ). This is impossible because there is no

way that a linear combination g(X, Y )X 2 + h(X, Y )Y 3 can produce X.

3. Since the localization functor is exact, we have (a) implies (b), and (b) implies (c) is

immediate. To prove that (c) implies (a), consider the exact sequence

i π

0 ’ ’ ’ im f ’ ’ ’ ker g ’ ’ ’ ker g/ im f ’ ’ ’ 0

’’ ’’ ’’ ’’

Applying the localization functor, we get the exact sequence

i π

0 ’ ’ ’ (im f )P ’ ’ ’ (ker g)P ’ ’ ’ (ker g/ im f )P ’ ’ ’ 0

’’ ’P’ ’P’ ’’

for every prime ideal P . But by basic properties of localization,

(ker g/ im f )P = (ker g)P /(im f )P = ker gP / im fP ,

which is 0 for every prime ideal P , by (c). By (1.5.1), ker g/ im f = 0, in other words,

ker g = im f , proving (a).

4. In the injective case, apply Problem 3 to the sequence

f

0 ’ ’ ’ M ’ ’ ’ N,

’’ ’’

3

and in the surjective case, apply Problem 3 to the sequence

f

M ’ ’ ’ N ’ ’ ’ 0.

’’ ’’

5. This follows because S ’1 (©Ai ) ⊆ ©i S ’1 (Ai ) for arbitrary rings (or modules) Ai .

6. Taking S = R \ Q and applying Problem 5, we have the following chain of inclusions,

where P ranges over all maximal ideals of R:

MQ = (©P RP )Q ⊆ ©P (RP )Q ⊆ (RQ )Q = RQ .

7. Since R is contained in every RP , we have R ⊆ M , hence RQ ⊆ MQ for every maximal

ideal Q. Let i : R ’ M and iQ : RQ ’ MQ be inclusion maps. By Problem 6,

RQ = MQ , in particular, iQ is surjective. Since Q is an arbitrary maximal ideal, i is

surjective by Problem 4, so R = M . But R ⊆ ©P prime RP ⊆ M , and the result follows.

8. The implication (a) implies (b) follows from (2.2.6), and (b) immediately implies (c).

To prove that (c) implies (a), note that if for every i, K is the fraction ¬eld of Ai , where

the Ai are domains that are integrally closed in K, then ©i Ai is integrally closed. It

follows from Problem 7 that R is the intersection of the RQ , each of which is integrally

closed (in the same fraction ¬eld K). Thus R is integrally closed.

9. The elements of the ¬rst ¬eld are a/f + P RP and the elements of the second ¬eld are

(a + P )/(f + P ), where in both cases, a, f ∈ R, f ∈ P . This tells you exactly how to

/

construct the desired isomorphism.

Chapter 3

1. Assume that (V, MV ) ¤ (R, MR ), and let ± be a nonzero element of R. Then either ±

or ±’1 belongs to V . If ± ∈ V we are ¬nished, so assume ± ∈ V , hence ±’1 ∈ V ⊆ R.

/

’1

Just as in the proof of Property 9 of Section 3.2, ± is not a unit of V . (If b ∈ V and

b±’1 = 1, then ± = ±±’1 b = b ∈ V .) Thus ±’1 ∈ MV = MR © V , so ±’1 is not a

unit of R. This is a contradiction, as ± and its inverse both belong to R.

2. 2. By de¬nition of h, ker h = MV . Since h1 extends h, ker h = (ker h1 ) © V , that is,

MV = MR1 © V . Since R1 ⊇ V , the result follows.

3. By hypothesis, (V, MV ) is maximal with respect to domination, so (V, MV ) = (R1 , MR1 ).

Therefore V = R1 , and the proof is complete.

4. If (R, MR ) is not dominated in this way, then it is a maximal element in the domination

ordering, hence R itself is a valuation ring.

Chapter 4

1. We have f ∈ I d i¬ all terms of f have degree at least d, so if we identify terms of degree

at least d + 1 with 0, we get an isomorphism between I d /I d+1 and the homogeneous

polynomials of degree d. Take the direct sum over all d ≥ 0 to get the desired result.

2. If x ∈ Mn and f (x) ∈ Nn+1 , then f (x) + Nn+1 = 0, so x ∈ Mn+1 .

4

3. The result holds for n = 0 because M0 = M and N0 = N . If it is true for n, let

x ∈ f ’1 (Nn+1 ). Since Nn+1 ⊆ Nn , it follows that x belongs to f ’1 (Nn ), which is

contained in Mn by the induction hypothesis. By Problem 2, the result is true for

n + 1.

4. Using the additional hypothesis and Problem 3, we have f ’1 (0) ⊆ f ’1 (©Nn ) =

©f ’1 (Nn ) ⊆ ©Mn = 0.

5. By (4.1.8) we have

(I m+k M ) © N = I k ((I m ) © N ) ⊆ I k N ⊆ (I k M ) © N.

6. Since gn —¦ fn = 0 for all n, we have g —¦ f = 0. If g(y) = 0, then y is represented

by a sequence {yn } with yn ∈ Mn and gn (yn ) = 0 for su¬ciently large n. Thus for

some xn ∈ Mn we have yn = fn (xn ). The elements xn determine x ∈ M such that

y = f (x), proving exactness.

7. Since R —R R ∼ R and tensor product commutes with direct sum, hM is an iso-

ˆ =ˆ

morphism when M is free of ¬nite rank. In general, we have an exact sequence

GN f GF GM G0

g

0

with F free of ¬nite rank. Thus the following diagram is commutative, with exact

G R— F G R— M G0

rows. ˆ ˆR ˆR

R —R N

hN hF hM

GN GF GM G0

ˆ ˆ ˆ

0 ˆ ˆ

g

f

See (4.2.7) for the last row. Since g is surjective and hF is an isomorphism, it follows

ˆ

that hM is surjective.

8. By hypothesis, N is ¬nitely generated, so by Problem 8, hN is surjective. Since hF is

an isomorphism, hM is injective by the four lemma. (See TBGY, 4.7.2, part (ii).)

9. Take inverse limits in (4.2.9).

10. Consider the diagram for Problem 7, with M ¬nitely generated. No generality is lost;

see TBGY, (10.8.1). Then all vertical maps are isomorphisms, so if we augment the

¬rst row by attaching 0 ’ on the left, the ¬rst row remains exact. Thus the functor

ˆ ˆ

R —R ” is exact, proving that R is ¬‚at.

ˆ

11. Since M is isomorphic to its completion, we may regard M as the set of constant

sequences in M . If x belongs to Mn for every n, then x converges to 0, hence x and

ˆ

0 are identi¬ed in M . By (4.2.4), the topology is Hausdor¬.

ˆ ˆ ˆ

12. I is ¬nitely generated, so by Problem 8, hI : R —R I ’ I is an isomorphism. Since R

is ¬‚at over R by Problem 10, R —R I ’ R —R R ∼ R is injective, and the image of

ˆ ˆ =ˆ

ˆ

this map is RI.

13. By Problem 12, (I n ) ∼ RI n = (RI)n ∼ (I)n .

ˆ= ˆ ˆ =ˆ

14. The following diagram is commutative, with exact rows.

5

G I n /I n+1 G R/I n+1 G R/I n G0

0

G (I)n /(I)n+1 G R/(I)n+1 G R/(I)n G0

ˆ ˆ ˆˆ ˆˆ

0

The second and third vertical maps are isomorphisms by (4.2.9), so the ¬rst vertical

map is an isomorphism by the short ¬ve lemma.

ˆ ˆ

15. By (4.2.9) and Problem 9, R is complete with respect to the I-adic topology. Suppose

ˆ ˆ

that a ∈ I. Since an + an+1 + · · · + am ∈ (I)n for all n, the series 1 + a + a2 + · · · + an

ˆ

converges to some b ∈ R. Now (1 ’ a)(1 + a + a2 + · · · + an ) = 1 ’ an+1 , and we can

ˆ ˆ

let n approach in¬nity to get (1 ’ a)b = 1. Thus a ∈ I ’ 1 ’ a is a unit in R. Since

ˆ ˆ ˆ

ax belongs to I for every x ∈ R, 1 + ax is also a unit. By (0.2.1), a ∈ J(R).

16. By (4.2.9), R/M ∼ R/M, so R/M is a ¬eld, hence M is a maximal ideal. By

=ˆ ˆ ˆˆ ˆ

ˆ ˆ

Problem 15, M is contained in every maximal ideal, and it follows that M is the

ˆ

unique maximal ideal of R.

Chapter 5

1. The function 2n is its own di¬erence.

2. If P is a prime ideal containing ann(M/MM ), then P ⊇ M, hence P = M by

maximality of M. Conversely, we must show that M ⊇ ann(M/MM ). This will be

true unless ann(M/MM ) = R. In this case, 1 annihilates M/MM , so MM = M . By

NAK, M = 0, contradicting the hypothesis.

3. Let S = R \ P . Then (R/I)P = 0 i¬ S ’1 (R/I) = 0 i¬ S ’1 R = S ’1 I i¬ 1 ∈ S ’1 I i¬

1 = a/s for some a ∈ I and s ∈ S i¬ I © S = 0 i¬ I is not a subset of P .

4. By Going Up [see (2.2.3)], any chain of distinct prime ideals of R can be lifted to a

chain of distinct prime ideals of S, so dim S ≥ dim R. A chain of distinct prime ideals

of S contracts to a chain of prime ideals of R, distinct by (2.2.1). Thus dim R ≥ dim S.

5. Since S/J is integral over the subring R/I, it follows from (5.3.1) and Problem 4 that

coht I = dim R/I = dim S/J = coht J.

6. If J is a prime ideal of S, then I = J © R is a prime ideal of R. The contraction of a

chain of prime ideals of S contained in J is a chain of prime ideals of R contained in

R, and distinctness is preserved by (2.2.1). Thus ht J ¤ ht I. Now let J be any ideal

of S, and let P be a prime ideal of R such that P ⊇ I and ht P = ht I. (If the height

of I is in¬nite, there is nothing to prove.) As in the previous problem, S/J is integral

over R/I, so by Lying Over [see (2.2.2)] there is a prime ideal Q containing J that lies

over P . Thus with the aid of the above proof for J prime, we have ht J ¤ ht Q ¤ ht

P = ht I.

7. First assume J is a prime ideal of S, hence I is a prime ideal of R. A descending chain

of distinct prime ideals of R starting from I can be lifted to a descending chain of

distinct prime ideals of S starting from J, by Going Down [see (2.3.4)]. Thus ht J ≥

ht I. For any ideal J, let Q be a prime ideal of S with Q ⊇ J. Then P = Q © R ⊇ I.

6

By what we have just proved, ht Q ≥ ht P , and ht P ≥ ht I by de¬nition of height.

Taking the in¬mum over Q, we have ht J ≥ ht I. By Problem 6, ht J = ht I.

8. The chain of prime ideals (X) ‚ (X, Y ) ‚ (X, Y , Z) gives dim R ≥ 2. Since XY (or

equally well XZ), belongs to the maximal ideal (X, Y, Z) and is not a zero-divisor, we

have dim R ¤ dim S/(XY ) = dim S ’ 1 = 2 by (5.4.7) and (5.4.9).

9. The height of P is 0 because the ideals (Y ) and (Z) are not prime. For example,

X ∈ (Y ) and Z ∈ (Y ), but X Z = 0 ∈ (Y ). Since R/P ∼ k[[X]] has dimension 1, P

/ / =

has coheight 1 by (5.3.1).

Chapter 6

1. By (6.1.3), dim R/P = dim R ’ t = dim R ’ ht P . By (5.3.1), dim R/P = coht P , and

the result follows.

2. Let J be the ideal (Z, X + Y ). If M = (X, Y, Z) is the unique maximal ideal of S,

2 2 2 2

then M = (X , Y , Z , Y Z) ⊆ J ⊆ M, so J is an ideal of de¬nition. (Note that

2 2

X Y = X Z = 0, X(X + Y ) = X , and Y (X + Y ) = Y .) By (6.1.2), {Z, X + Y } is

a system of parameters. Since Z X = 0, Z is a zero-divisor.

Chapter 7

1. Note that ker f, im f , and ker g are all equal to {0, 2}.

2. We have im ‚n = ker fn’1 = 0 and ker gn = im fn = Bn . Thus gn is the zero map, so

ker ‚n = im gn = 0. Therefore ‚n is an injective zero map, which forces Cn = 0.

3. This follows from the base change formula R/I —R M ∼ M/IM with I = M (see

=

TBGY, S7.1).

4. We have g — : 1 — ei ’ 1 — xi , which is an isomorphism. (The inverse is 1 — xi ’ 1 — ei .)

Thus im f — = ker g — = 0. Since f — is the zero map, δ is surjective. But ker uM is 0 by

hypothesis, so δ = 0. This forces coker uK = 0.

5. By Problem 4, K = MK. Since M is a Noetherian R-module, K is ¬nitely generated,

so by NAK we have K = 0. Thus 0 = im f = ker g, so g is injective.

6. Since free implies projective implies ¬‚at always, it su¬ces to show that ¬‚at implies

free. If M is ¬‚at, then the functor N ’ N —R M is exact. If M is the maximal ideal

of R, then the map M —R M ’ R —R M ∼ M via a — x ’ ax is injective. But this

=

map is just uM , and the result follows from Problems 3-5.

7. We have the short exact sequence 0 ’ I ’ R ’ R/I ’ 0, which induces, for any

R-module N , the exact sequence

HomR (R/I, N ) ’ HomR (R, N ) ’ HomR (I, N ) ’ Ext1 (R/I, N ).

R

The last term is 0 by hypothesis, hence the map i— : HomR (R, N ) ’ HomR (I, N ) is

surjective. This says, by Baer™s criterion (TBGY 10.6.4), that N is injective.

7

8. The left side is at least equal to the right side, so assuming that the right side is at

most n, it su¬ces to show that idR N ¤ n for all N . Given an exact sequence as

in (7.2.4) part 4, dimension shifting yields Extn+1 (R/I, N ) ∼ Ext1 (R/I, Cn’1 ). By

= R

R

1

(7.1.7), ExtR (R/I, Cn’1 ) = 0, so by (7.2.1) and Problem 7, Cn’1 is injective. By

(7.2.4), idR N ¤ n.

9. If (a) holds, only the second assertion of (b) requires proof. Apply Tor to the exact

sequence 0 ’ N ’ N ’ N ’ 0 to get the exact sequence

0 = TorR (M, N ) ’ M —R N ’ M —R N ’ M —R N ’ 0.

1

We may replace M —R N by (M —R S) —S N , and similarly for the other two tensor

products. By exactness, M —R S is ¬‚at. Now assuming (b), we have TorR (M, F ) = 0

1

for every free S-module F , because Tor commutes with direct sums. If N is an arbitrary

S-module, we have a short exact sequence 0 ’ K ’ F ’ N ’ 0 with F free. The

corresponding (truncated) long exact sequence is

0 = TorR (M, F ) ’ TorR (M, N ) ’ M —R K ’ M —R F ’ M —R N ’ 0.

1 1

As before, we replace M —R K by (M —R S)—S K, and similarly for the other two tensor

products. The map whose domain is (M —R S) —S K is induced by the inclusion of K

into F , and is therefore injective, because M —R S is a ¬‚at S-module by hypothesis.

Thus the kernel of the map, namely TorR (M, N ), is zero.

1

Chapter 8

1. To ease the notation we will omit all the overbars and adopt the convention that all

calculations are mod (X 3 ’ Y 2 ). We have (X 2 + X + 1)(X ’ 1) = X 3 ’ 1 = Y 2 ’ 1 =

(Y ’ 1)(Y + 1). Now X 2 + X + 1 and Y + 1 are units in R because they do not vanish

when X = Y = 1, assuming that the characteristic of K is not 2 or 3. Thus X ’ 1 and

Y ’ 1 are associates.

2. The maximal ideal is not principal because X and Y cannot both be multiples of a single

polynomial. To show that dim R = 1, we use (5.6.7). Since K(Y ) has transcendence

degree 1 over K and K(X, Y )/(X 3 ’ Y 2 ) is algebraic over K(Y ), (we are adjoining

a root of X 3 ’ Y 2 ), it follows that the dimension of K[X, Y ]/(X 3 ’ Y 2 ) is 1. By

(5.3.1), the coheight of (X 3 ’ Y 2 ) is 1, and the corresponding sequence of prime

ideals is (X 3 ’ Y 2 ), (X, Y ). Thus localization at (X, Y ) has no e¬ect on dimension, so

dim R = 1. (In general, prime ideals of a localized ring AP correspond to prime ideals

of A that are contained in P , so localization may reduce the dimension.)

3. By de¬nition, the Hilbert polynomial is the composition length lk (I n /I n+1 ). Since

monomials of degree n in r variables form a basis for the polynomials of degree n, we

must count the number of such monomials, which is

n+r’1 (n + r ’ 1)(n + r ’ 2) · · · (n + 2)(n + 1)

=

r’1 (r ’ 1)!

This is a polynomial of degree r ’ 1 in the variable n.

8

4. This follows from Problem 3 and additivity of length (5.2.3).

5. Fix a nonzero element b ∈ Bd . (Frequently, b is referred to as a homogeneous element

of degree d.) By de¬nition of a graded ring, we have bAn ⊆ Bn+d for n ≥ 0. Then

lk (Bn+d ) ≥ lk (bAn ) = lk (An ) ≥ lk (Bn ).

n+r’1

Since lk (An ) = , the result follows.

r’1

6. If R is regular, we may de¬ne the graded k-algebra homomorphism • of Problems

3-5 with r = d. Since the Hilbert polynomial has degree d, • is an isomorphism.

Conversely, an isomorphism of graded k-algebras induces an isomorphism of ¬rst com-

ponents, in other words,

(k[X1 , . . . , Xd ])1 ∼ M/M2 .

=

But the k-vector space on the left has a basis consisting of all monomials of degree 1.

Since there are exactly d of these, we have dimk M/M2 = d. By (8.1.3), R is regular.